why does byte change when bit shifting?

Question

Hi everyone. I'm playing with bit shifting. I'm trying to take a 32bit int, save each byte in an array of char, then reconstitute the int. It works the way I think it should, except the second byte from the right seems to have the lowest bit switched. My code is:

int main() {
  char paus[2];
  char b[4] = "abc";
  int c = 6104;
  int d = 0xcccccccc;
  printf("c in hex: %x\n",c);
  printf("d in hex: %x\n",d);
  printf("b[0]: %x\nb[1]: %x\n",b[0]&0xff,b[1]&0xff);
  printf("b[2]: %x\nb[3]: %x/n",b[2]&0xff,b[3]&0xff);
  printf("\n");

  b[0] = c >> 24;
  b[1] = (c >> 16) & 0xff;
  b[2] = (c >> 8) & 0xff;
  b[3] = c & 0xff;
  printf("b[0]: %x\nb[1]: %x\n",b[0]&0xff,b[1]&0xff);
  printf("b[2]: %x\nb[3]: %x\n",b[2]&0xff,b[3]&0xff);
  printf("\n");

  d = (d << 8) + 0x15;
  printf("d in hex: %x\n",d);
  d = (d << 8) + b[1];
  printf("d in hex: %x\n",d);
  d = (d << 8) + b[2];
  printf("d in hex: %x\n",d);
  d = (d << 8) + b[3];
  printf("d in hex: %x\n",d);

  fgets(paus,2,stdin);
  return 0;
}

The output is:

c in hex: 17d8
d in hex: cccccccc
b[0]: 61
b[1]: 62
b[2]: 63
b[3]: 0

b[0]: 0
b[1]: 0
b[2]: 17
b[3]: d8

d in hex: cccccc15
d in hex: cccc1500
d in hex: cc150017
d in hex: 150016d8

Everything makes sense except why the second byte from the right changes to 16 from bit shifting 17 left 8 bits? The 15 and 00 bytes are carried all the way, so why does the 17 byte change? Thanks!

Answer 1

The problem is that your compiler is set to treat char as signed char. Try this:

d = (d << 8) + (unsigned char)b[3];
printf("d in hex: %x\n",d);

(You would probably want to do the same for the other instances.)

Alternatively, you can change the declaration of d:

unsigned char b[4] = "abc";